| Date | May 2015 | Marks available | 3 | Reference code | 15M.1.hl.TZ2.12 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Show that | Question number | 12 | Adapted from | N/A |
Question
The cubic equation \({x^3} + p{x^2} + qx + c = 0\), has roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \). By expanding \((x - \alpha )(x - \beta )(x - \gamma )\) show that
(i) \(p = - (\alpha + \beta + \gamma )\);
(ii) \(q = \alpha \beta + \beta \gamma + \gamma \alpha \);
(iii) \(c = - \alpha \beta \gamma \).
It is now given that \(p = - 6\) and \(q = 18\) for parts (b) and (c) below.
(i) In the case that the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) form an arithmetic sequence, show that one of the roots is \(2\).
(ii) Hence determine the value of \(c\).
In another case the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) form a geometric sequence. Determine the value of \(c\).
Markscheme
(i)-(iii) given the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \), we have
\({x^3} + p{x^2} + qx + c = (x - \alpha )(x - \beta )(x - \gamma )\) M1
\( = \left( {{x^2} - (\alpha + \beta )x + \alpha \beta } \right)(x - \gamma )\) A1
\( = {x^3} - (\alpha + \beta + \gamma ){x^2} + (\alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma \) A1
comparing coefficients:
\(p = - (\alpha + \beta + \gamma )\) AG
\(q = (\alpha \beta + \beta \gamma + \gamma \alpha )\) AG
\(c = - \alpha \beta \gamma \) AG
[3 marks]
METHOD 1
(i) Given \( - \alpha - \beta - \gamma = - 6\)
And \(\alpha \beta + \beta \gamma + \gamma \alpha = 18\)
Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \)
So \(\beta - \alpha = \gamma - \beta \) M1
or \(2\beta = \alpha + \gamma \)
Attempt to solve simultaneous equations: M1
\(\beta + 2\beta = 6\) A1
\(\beta = 2\) AG
(ii) \(\alpha + \gamma = 4\)
\(2\alpha + 2\gamma + \alpha \gamma = 18\)
\( \Rightarrow {\gamma ^2} - 4\gamma + 10 = 0\)
\( \Rightarrow \gamma = \frac{{4 \pm {\text{i}}\sqrt {24} }}{2}\) (A1)
Therefore \(c = - \alpha \beta \gamma = - \left( {\frac{{4 + {\text{i}}\sqrt {24} }}{2}} \right)\left( {\frac{{4 - {\text{i}}\sqrt {24} }}{2}} \right)2 = - 20\) A1
METHOD 2
(i) let the three roots be \(\alpha ,{\text{ }}\alpha - d,{\text{ }}\alpha + d\) M1
adding roots M1
to give \(3\alpha = 6\) A1
\(\alpha = 2\) AG
(ii) \(\alpha \) is a root, so \({2^3} - 6 \times {2^2} + 18 \times 2 + c = 0\) M1
\(8 - 24 + 36 + c = 0\)
\(c = - 20\) A1
METHOD 3
(i) let the three roots be \(\alpha ,{\text{ }}\alpha - d,{\text{ }}\alpha + d\) M1
adding roots M1
to give \(3\alpha = 6\) A1
\(\alpha = 2\) AG
(ii) \(q = 18 = 2(2 - d) + (2 - d)(2 + d) + 2(2 + d)\) M1
\({d^2} = - 6 \Rightarrow d = \sqrt 6 {\text{i}}\)
\( \Rightarrow c = - 20\) A1
[5 marks]
METHOD 1
Given \( - \alpha - \beta - \gamma = - 6\)
And \(\alpha \beta + \beta \gamma + \gamma \alpha = 18\)
Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \).
So \(\frac{\beta }{\alpha } = \frac{\gamma }{\beta }\) M1
or \({\beta ^2} = \alpha \gamma \)
Attempt to solve simultaneous equations: M1
\(\alpha \beta + \gamma \beta + {\beta ^2} = 18\)
\(\beta (\alpha + \beta + \gamma ) = 18\)
\(6\beta = 18\)
\(\beta = 3\) A1
\(\alpha + \gamma = 3,{\text{ }}\alpha = \frac{9}{\gamma }\)
\( \Rightarrow {\gamma ^2} - 3\gamma + 9 = 0\)
\( \Rightarrow \gamma = \frac{{3 \pm {\text{i}}\sqrt {27} }}{2}\) (A1)(A1)
Therefore \(c = - \alpha \beta \gamma = - \left( {\frac{{3 + {\text{i}}\sqrt {27} }}{2}} \right)\left( {\frac{{3 - {\text{i}}\sqrt {27} }}{2}} \right)3 = - 27\) A1
METHOD 2
let the three roots be \(a,{\text{ }}ar,{\text{ }}a{r^2}\) M1
attempt at substitution of \(a,{\text{ }}ar,{\text{ }}a{r^2}\) and \(p\) and \(q\) into equations from (a) M1
\(6 = a + ar + a{r^2}\left( { = a(1 + r + {r^2})} \right)\) A1
\(18 = {a^2}r + {a^2}{r^3} + {a^2}{r^2}\left( { = {a^2}r(1 + r + {r^2})} \right)\) A1
therefore \(3 = ar\) A1
therefore \(c = - {a^3}{r^3} = - {3^3} = - 27\) A1
[6 marks]
Total [14 marks]
